Transcript Cct component - Universiti Sains Malaysia

Transformer -2
(a) Iron-cored
Transformer
(d)
Tapped windings B
A
(b) Ferrite –cored
Transformer
(e) AutoTransformer
(c)
Multiplewindings
F
E
D
C
C
B
H
B
A
G
F
E
D
C
A
Construction of transformer
Laminated steel-core transformer
Primary windings of a 30 kVA, 6000 V/230 V transformer has
a resistance of 10 , the secondary windings has a resistance
of 0.016 . The total reactance of the transformer referred to
primary is 23 . Calculate the voltage regulation of the
transformer when it supplies a the full-load current at power
factor of 0.8 lagging
I1
R1
10 
V1
6000/230 V
Xe
23 
V1’
R2
I2
0.016

V2’
30 kVA
V2
Equivalent resistance of primary and secondary referred to primary is
2
 V1 
 6000
Re  R1  R2    10  0.016
  21
V
230


 2
2
cos2  0.8
I1 Re
sin 2  0.6
Full-load current at primary
6000/230
Xe
I2
V
V1
V1
V2
’
’
30
KVA 30000
I1 

 5A
V
6000
kVA
Per  unit voltageregulation
I1 Re cos2  X e sin 2 
V1
521  0.8  23  0.6 

 0.0255
6000
 2.55 %
V2
Losses in transformer on load categorized into two
1. I2R losses in primary and secondary windings,
namely I12R1+I22R2.
2. Core losses due to hysteresis and eddy currents
Usually variation between no-load and full-load can be
negligible thus, the total core loss, PC , is assumed to be
constant all the time.
2
2

P

I
R

I
Therefore total loss
C
1 1
2 R2
or
 PC  I12 R1e  PC  I 22 R2e
R1e=equivalent resistance of primary and secondary referred to primary
R2e=equivalent resistance of primary and secondary referred to secondary
output power
output power
Efficiency

input power output power  losses
I 2V2  p. f .

I 2V2  p. f .  Pc  I12 R1  I 22 R2
Note: p.f= power factor
The wire losses can be expressed as
Efficiency
Divided by I2
V
R2e  R2  R1  2
 V1
2

N
  R2  R1  2

 N1
I 2V2 cos.
I 2V2 cos.  Pc  I 22 R2e
V2 cos.
Efficiency
V2 cos.  Pc / I 2  I 2 R2e



2
It is also possible to express
output power input power  losses
Efficiency

input power
input power
Or
losses
  1
input power
The primary and secondary windings of a 500 kVA transformer
have a resistances of 0.42  and 0.0011  respectively. The
primary and secondary voltages are 6600 V and 400 V
respectively and the core loss is 2.9 kW , assuming the power
factor of the load to be 0.8. Calculate the efficiency on (a) fullload and (b) half-load. (c) assuming the power factor 0.8, find
output which the efficiency of the transformer is maximum.
(a)
Primary current on full-load
Secondary current on full-load
500103
I1 
 75.8 A
6600
500103
I2 
 1250A
400
Coil Wire loss at primary PW1=I12R1=75.82 x 0.42=2415W
Coil Wire loss at secondary PW2=I22R2=12502 x 0.0011=1720W
PW = PW1 + PW2 = 2415 + 1720 = 4.135kW
Total loss PL= PW + PC = 4.135 + 2.9 = 7.035kW
Output power on full load Pout = 500 x 0.8 = 400 kW
Therefore Pin = Pout + PL = 400+7.035 = 407.035 kW
losses
7.035
  1
 1
 0.9827 per unit
Input
407.035
 98.27%
(b)
Since the wire loss varies as square of the current , thus
Losses on half-load PW/2=4.135/22=4.135/4=1.034kW
Total Loss on half-load PL= PC+PW/2=2.9+1.034=3.934kW
Output power on half-load Pout/2= 400/2 = 200kW
Input power on half-load Pin/2= Pout/2+PL=200+3.934 = 203.934kW
Loss
3.934
  1
 1
 0.9807 per unit
Input
203.934
 98.07%
(c)
Full-load I2R loss is PW = 4.135kW
Let n= fraction of full-load appearance power at
which it is maximum efficiency
Total I2R loss is = n2 x 4.135 kW=2.9
Therefore n=0.837
Output at maximum efficiency is= 0.837 x 500= 418.5kWA
Output power at power factor 0.8= 418.5 x 0.8 = 334.8 kWA
Since the core and I2R are equaled, then total loss is
PL= 2 x 2.9 = 5.8 kW
Maximum   1 
Loss
5.8
 1
 0.983 per unit  98.3%
Input
334.8  5.8